How do you simplify #(root6(8)*root6(16))/root6(2)#?

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Answer 1 · 2016-08-26T15:25:24

#2#

Look for values you can cancel out:

write as:#" "(root(6)(cancel(2)xx4)xxroot(6)(16))/(cancel(root(6)(2)))#

#root(6)(4xx16)#

#root(6)(64)=2#

Answer 2 · 2016-08-29T08:47:05

#2#

When we are multiplying or dividing with roots, we can combine them into one root. (as long as it is the same root in each case.)

#(root6(8) xxroot6(16))/root6(2)#

=#root6((8xx16)/2)#

Simplifying gives: #root6(64)#

(It is very useful to know the powers of 2 up to #2^10 = 1024#)

#root6(64) = root6(2^6)#

=#2#